The Pellian Equation x2 −Dy2 = 1 A General Method (Part XIIB)

A Method for Generating Pellian Triples from Paired Sequences P(n) (x,y,1)

The Pellian equation is the Diophantine equation x2 − Dy2 = z2 where z equals 1. The least solutions of the Pell equation are posted in Wikipedia. and also listed in Table 91, page 254 of Recreations in the Theory of Numbers by Albert H. Beiler (1966), where the values for D on page 252-253 have been computed using the following two expressions:

x = [(p + q√D)n + (p − q√D)n ∕ 2]
y = [(p + q√D)n + (p − q√2D)n ∕ 2√2D)]

The tables in these two articles show a series of numbers of which I will focus on those triples of the type (x,y,1) starting at (3,2,1) and D = 2. It will be shown here that we can continue generating all those values of x not listed in these articles by employing a general sequence method where the specific seqeunces from this general method may or may not be in the OEIS database. The sequences uses the expression consisting of a pair of numbers and the corresponding x values while keeping y constant:

P(n) = n(an − 2), n(an + 2)
x(n) = an − 1, an + 1

corresponding to the various D and x values in the sequence where each pair uses the same value of n, and a is equal to the square of y. Thus, this method is a way of obtaining the Pellian triples from the set of square numbers S = {(y1)2,(y2)2,(y3)2,...} of which a miniscule number are found in the tables of the above articles.

However, as a check once we find D and its corresponding x value we can backtrack and find y according to the following:

y = ⌊ x/√D ⌋ = a½

where y is the floor(y) value, i.e., rounded down to the nearest integer equal to the square root of a.

Accordingly, Tables D and E shows the calculation of P(n) and x(n) for a, viz., the square numbers 1,4,9,16,25,36,49,64,81 and n 1 to 5, producing the following tabulated results for D and x. Values of y from the floor(y) are calculated from the previous two rows, another way of verifying for the value of y. Moreover, when y = 1 the D and xs are repeated so that we can eliminate the n(n − 2) and n − 1 and just focus on the n(n + 2) and n + 1 . The sequence for these D numbers (where y = 1 in Table D) is listed in the OEIS database entry A005563 as n(n + 2).

Table D
n 11223344 55
D-13083158241535
x0213243546
y1111111111
D2612203042567290110
x3579111315171921
y2222222222
D71132407587136152215235
x8101719262835375355
y3333333333
D14186068138150248264390410
x15173133474963657981
y4444444444
D232796104219231392408615635
x24264951747699101124126
y5555555555
Table E
n 11223344 55
D3438140148318330568584890910
x35377173107109143145179181
y6666666666
D475119220043544771679212151235
x48509799146148195197244246
y7777777777
D62662522605705821016103215901610
x6365127129191193255257319321
y8888888888
D79833203287237351288130420152035
x8082161163242244323325404406
y9999999999

The method of generating groups of triples involves multiplying the initial least solutions by either of the two parts of the following mathematical expression:

RD = (n1y + √D)2 ∕ 2n1 = x + y√D)

where:

n1 are consecutive integers, 1, 2, 3, 4, 5, ...
the values of adjacent xs in a pair is (an1 − 1, an1 + 1).
D are the values from the above sequences starting at column one.

then multiplying and rounding off each row of triples generated by the RD for as many triples as are desired. The lists below show the patterns generated for a subset of Ds from tables D and E (starting at the yellow 14) and the accompanying dual equations for each D from which the triple tables, similar to those of Part XIIA, may be constructed.

D and Corresponding Pell (x,y,1) Triples

Table 4,5
The Pell equation for each example is x2 − 4y2 = 1 The Pell equation for each example is x5 − 5y2 = 1.
R14 = (4 + √14)2 ∕ 2 = 15 + 4√14 R23 = (5 + √23)2 ∕ 2 = 24 + 5√23
R18 = (4 + √18)2 ∕ 2 = 17 + 4√18 R27 = (5 + √27)2 ∕ 2 = 36 + 5√27
R60 = (8 + √60)2 ∕ 4 = 31 + 4√60 R96 = (10 + √96)2 ∕ 4 = 49 + 5√96
R68 = (8 + √68)2 ∕ 4 = 33 + 4√68 R104 = (10 + √104)2 ∕ 4 = 51 + 5√104
R138 = (12 + √138)2 ∕ 6 = 47 + 4√138 R219 = (15 + √219)2 ∕ 6 = 74 + 5√219
R150 = (12 + √150)2 ∕ 6 = 49 + 4√150 R231 = (15 + √231)2 ∕ 6 = 76 + 5√231
Table 6,7
The Pell equation for each example is x2 − 6y2 = 1 The Pell equation for each example is x2 − 7y2 = 1.
R34 = (6 + √34)2 ∕ 2 = 35 + 6√34 R47 = (7 + √47)2 ∕ 2 = 48 + 7√47
R38 = (6 + √38)2 ∕ 2 = 37 + 6√38 R51 = (7 + √51)2 ∕ 2 = 50 + 7√51
R140 = (12 + √140)2 ∕ 4 = 71 + 6√140 R192 = (14 + √192)2 ∕ 4 = 97 + 7√192
R148 = (12 + √148)2 ∕ 4 = 73 + 6√148 R200 = (14 + √200)2 ∕ 4 = 99 + 7√200
R318 = (18 + √318)2 ∕ 6 = 107 + 6√318 R435 = (21 + √435)2 ∕ 6 =146 + 7√435
R330 = (18 + √330)2 ∕ 6 = 109 + 6√330 R447 = (21 + √447)2 ∕ 6 = 148 + 7√447
Table 8,9
The Pell equation for each example is x2 − 8y2 = 1 The Pell equation for each example is x2 − 9y2 = 1.
R62 = (8 + √62)2 ∕ 2 = 63 + 8√62 R79 = (9 + √79)2 ∕ 2 = 80 + 9√79
R66 = (8 + √66)2 ∕ 2 = 65 + 8√66 R83 = (9 + √83)2 ∕ 2 = 82 + 9√83
R252 = (16 + √252)2 ∕ 4 = 127 + 8√252 R320 = (18 + √320)2 ∕ 4 = 161 + 9√320
R260 = (16 + √260)2 ∕ 4 = 129 + 8√260 R328 = (18 + √328)2 ∕ 4 = 163 + 9√328
R570 = (24 + √570)2 ∕ 6 = 191 + 8√570 R723 = (27 + √723)2 ∕ 6 =242 + 9√723
R582 = (24 + √582)2 ∕ 6 = 193 + 8√582 R735 = (27 + √735)2 ∕ 6 = 244 + 9√735

This concludes Part XIIB. Go back to Part XIIA.

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Copyright © 2020 by Eddie N Gutierrez. E-Mail: enaguti1949@gmail.com